If a1,a2,a3,…,a24 are  in arithmetic progression  and  a1+a5+a10+a15+a20+a24=225 then a1+a2+a3+…+a23+a24 is equal to

# If ${\mathrm{a}}_{1},{\mathrm{a}}_{2},{\mathrm{a}}_{3},\dots ,{\mathrm{a}}_{24}$ are  in arithmetic progression  and  ${\mathrm{a}}_{1}+{\mathrm{a}}_{5}+{\mathrm{a}}_{10}+{\mathrm{a}}_{15}+{\mathrm{a}}_{20}+{\mathrm{a}}_{24}=225$ then ${\mathrm{a}}_{1}+{\mathrm{a}}_{2}+{\mathrm{a}}_{3}+\dots +{\mathrm{a}}_{23}+{\mathrm{a}}_{24}$ is equal to

1. A

909

2. B

75

3. C

750

4. D

900

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### Solution:

Given that,

${\mathrm{a}}_{1}+{\mathrm{a}}_{5}+{\mathrm{a}}_{10}+{\mathrm{a}}_{15}+{\mathrm{a}}_{20}+{\mathrm{a}}_{24}=225$

$\begin{array}{l}⇒ \left({\mathrm{a}}_{1}+{\mathrm{a}}_{24}\right)+\left({\mathrm{a}}_{5}+{\mathrm{a}}_{20}\right)+\left({\mathrm{a}}_{10}+{\mathrm{a}}_{15}\right)=225\\ ⇒ 3\left({\mathrm{a}}_{1}+{\mathrm{a}}_{24}\right)=225\\ ⇒ {\mathrm{a}}_{1}+{\mathrm{a}}_{24}=75\end{array}$

[$\because$in an AP the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of the first and last term]

[from eq.(i)]  Register to Get Free Mock Test and Study Material

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